Extension of a map $g:\overline{B_1^n}\to \mathbb{R}^2$

Question

Let $B_r^n\subset \mathbb{R}^n$ ($n\geq 6$) be the open ball with radius $r$ and let $g:B_2^n\to \mathbb{R}^2$ be an analytic map. How to define a continuous map $h:\mathbb{R}^n\to \mathbb{R}^2$ such that $h|_{\overline{B_1^n}}=g|_{\overline{B_1^n}}$ and $h^{-1}(0)\subset \overline{B_1^n}$? If $P$ is a 2-dimensional vector subspace in $\mathbb{R}^n$ such that $(g|_P)^{-1}(0)=\{0\}$, then how to define a continuous map $h:\mathbb{R}^n\to \mathbb{R}^2$ such that $h|_{\overline{B_1^n}\cup P}=g|_{\overline{B_1^n}\cup P}$ and $h$ does not vanish outside of $B_2^n$?

Answer 1

It is not possible in general. For the first question:

If such function $h$ existed, it would mean that arbitrary small continuous perturbations $$\tilde{g}_\epsilon(x):=g((1+\epsilon) x)$$ of $g(x)$ exist such that $\tilde{g}_\epsilon$ is nowhere zero on $S^{n-1}=\partial B_1^n$. However, this is not possible if $g(x)$ has a "robust" zero on $S^{n-1}$. For example, take $$ g(x_1,\ldots, x_n)=(x_1, x_2). $$ Let $N:=(0,\ldots,0,1)\in S^{n-1}$, $U\simeq B^{n-1}$ be a neighborhood of $N$ in $S^{n-1}$, and $U':=U\cap \langle e_1, e_2, e_n\rangle$ be a two-dimensional "sub-neighborhood". Let $g':=g|_{U'}$. The degree of $g'/|g'|$ as a map from $\partial U'$ to $S^1$ has degree 1, so if $\tilde{g'}$ is a small enough perturbation (in the $L_\infty$ norm) of $g'$, then the degree of $\tilde{g'}/|\tilde{g'}|: \partial U'\to S^1$ will still be 1, so $\tilde{g'}$ can not be extended to $U'\to\mathbb{R}^2\setminus \{0\}$ and $\tilde{g'}=0$ has a solution in $U'$.

More generally, your extension problem has a solution iff for any $\epsilon>0$, the map $|g|^{-1}(\epsilon)\cap S^{n-1}\to S^1$ given by $g/|g|$ can be extended to some continuous map $|g|^{-1}[0,\epsilon]\cap S^{n-1}\to S^1$.

EDIT: Concerning your second question, the answwer is negative as well. You want to extend the map $g|_{P\cap S_2^{n-1}}$ that is nowhere zero on $P\cap S_2^{n-1}$ to a nowhere zero map on the whole $S_2^{n-1}$ (I denote by $S_2^{n-1}$ the boundary of $B_2^{n}$). But again, if the degree of your map $g/|g|: P\cap S_2^{n-1}\to S_1$ is nonzero, then you can not extend it to a map $S_2^{n-1}\to S^1$. You can not even extend it to the 2-skeleton of some triangulation of $S_2^{n-1}$.

An easy visualization is the case $n=3$, $g(x_1,x_2, x_3)=(x_1, x_2)$ and $P$ the span of $e_1, e_2$. Then any extension $S_2^2\to\mathbb{R}^2$ of $g|_{P\cap S_2^2}$ will contain at least one zero in the upper and at least one zero in the lower hemisphere.